Given reaction 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g), you react 5.0 L of NH3 with 5.0 L of Cl2 measured at the same conditions in a closed container. Calculate the ratio of pressures in the container (Pfinal/Pinitial).

Question

0.75
1.00
1.50
1.33
None of these choices are correct

Answer 1

2NH3 + 3Cl2 ==> N2 + 6HCl
Determine the limiting reagent. It is Cl2. Repost if you don't know how to do that. So all of the Cl2 will be used up.
How much N2 is formed?
5L x 1/3 = 1.67
How much HCl is formed?
5L x6/3 = 10 L
How much NH3 is used?
5L x 2/3 = 1.67 so how much NH3 is left? 5.0-1.67 = 3.33
So total L = 3.33 + 0 + 1.67 + 10 = 13.34. I have used L in all cases and that substitutes for mols in gas problems.
Total P initial = 5+5 = 10
Total P final = 13.34
You can do the ratio.

Answer 2

identify with resason oxidizing 3cuo (5) *2nh3 (g) 3cu (5)*3 h20 (1)*n2 (g)

Answer 3

n